Question
Let $f:[0,\infty)\times [0,1]\to\mathbb{R}$ be a continuous function. Show that $F(t)=\int_{0}^{1} f(t,x) \, dx$ is continuous on $[0,\infty)$.
Ideas
It feels like I'll need to apply the Dominated Convergence Theorem. A proof would probably have the shape:
Let $(t_n)$ be any sequence in $[0,\infty)$ converging to $t\in [0,\infty)$. Define $f_n(x)=f(t_n,x)$. As $f$ is continuous, we have $\lim_{n\to\infty} f_n(x)=f(t,x)$. [Now insert bit about dominating function $g$]. So
$F(t_n)=\int_{0}^{1} f(t_n,x) \, dx\to\int_{0}^{1} f(t,x) \, dx=F(t)$ by DCT. Hence $F$ is continuous on $[0,\infty)$.
But I'm stuck on finding a dominating function since $f$ is general on that domain. Also before we show $F$ is continuous, don't we need to say why it exists in the first place? How do I justify that from the information in the premises?
Thanks